Question

What am I doing wrong here with passing a char array to a function and in the function giving each index memory (using malloc()), then inserting something from the keyboard using gets().

#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <string.h>
#include <sys/wait.h>

void test(char *arr[]);
int main(){
  char *arr[2];//2 is the rows
 /* arr[0] = malloc(80);//This commented code works
  arr[1] = malloc(80);
  strcpy(arr[0], "hey");
  strcpy(arr[1], "whats up");
*/

  test(*arr);
  printf("in array[0]: %s", arr[0]);
  printf("in array[1]: %s", arr[1]);
  return 0;
}
void test(char *arr[]){
  int index;
  char *input = malloc(80);
  for(index = 0; index < 2; index++){
  arr[index] = malloc(80);
  gets(input);
  strcpy(arr[index], input);
  //arr[0] = input;
  }
}

Just a very basic program which for some reason I am having trouble with. Also one more question When I declare an array what is the difference between these forms

char *array

as oppose to

char *array[size]

or

char **array

Thanks,
Kevin

Answer 1

You declare arr as char *arr[2]. You then pass in *arr, which has type char* to test. But test expects a char *[]. So that won’t work. You should simply pass in arras it is, i.e. test(arr).

char * array is a pointer to character, typically used to point to the first character in an array of characters, i.e. a string.
char **array is a pointer to pointer to character. Typically used to represent an array of strings.
char *array[size] is mostly equivalent to the one above, but it unlike, that one, the top-level pointer is already pointing at a valid array, so the array does not need to be malloced.

By the way, your test function could be simplified a bit: the strcopy is unneccessary.

void test(char *arr[])
{
    int i;
    for(i=0;i<2;i++)
    {
        arr[i] = malloc(80);
        gets(arr[i]);
    }
}